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Power of DOE

Power of DOE. Some simple examples using 2-level factorials. Newton’s Law of Motion. Consider Newton’s Law of motion: F = m.a If nature knows this relationship but we don’t, how can we obtain the relationship without years of theoretical research?

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Power of DOE

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  1. Power of DOE Some simple examples using 2-level factorials

  2. Newton’s Law of Motion • Consider Newton’s Law of motion: F = m.a • If nature knows this relationship but we don’t, how can we obtain the relationship without years of theoretical research? • Let’s try to do it by experiment using the 2 factors mass (m) and acceleration (a) as input factors. • The response is the Force (F).

  3. Experimental Design • 2-level, 2 factors or 22 full factorial design • Inputs: M: 5, 10; A: 100, 200 Run M A Avg. Force 1 5 100 500 2 5 200 1000 3 10 100 1000 4 10 200 2000

  4. Analysis • Using a sign table: Run m a m x a Force 1 -1 -1 1 500 2 -1 1 -1 1000 3 1 -1 -1 1000 4 1 1 1 2000 Overall average Force = 1125 [m] = (1500)-(750) = 750, [a] = (1500) – (750) = 750 [m.a] = (1250) – (1000) = 250

  5. Regression Equation • F = 1125 + 375 m + 375 a + 125 m.a • Now we need to convert the coded values into actual values • M = (m+ + m-)/2 + (m+ - m-)/2 x m • i.e. M = 7.5 + 2.5 x m • Or m = (M -7.5)/2.5 • Likewise for the other terms

  6. Substituting the actual transformed values into the regression equation, • F = 1125 + 375 (M-7.5)/2.5 + 375 (A-150)/50 • + 125 (M-7.5)/2.5 x (A-150)/50 • Simplifying, F = M.A Imagine how quickly Newton would have closed the knowledge gap if he knew DOE!

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